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In other words:

For every natural number $m$, does there always exist an $n$ for which there are exactly $m$ groups of order $n$ up to isomorphism?

Or is this an open question in mathematics? If it is an open question, are there any famous conjectures one way or the other? And what progress has been made in answering the question?

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    $\begingroup$ Which partial results do you know? $\endgroup$
    – lhf
    Jun 21, 2018 at 17:12
  • $\begingroup$ The OEIS might be helpful. I was able to find up to $m = 21$ at the list here. $\endgroup$ Jun 21, 2018 at 17:13
  • $\begingroup$ @lhf Absolutely none; I came up with this conjecture on my own and wasn't able to find any relevant papers on it through internet searches. I did however find this list of the number of groups of order n through n=2015: orion.math.iastate.edu/maddux/504-Fall-2009/groups.pdf $\endgroup$
    – Izzhov
    Jun 21, 2018 at 17:14
  • $\begingroup$ It is also a well-known open problem whether there exists a number $n$ such that the number of isomorphism classes of groups of order $n$ is $n$. $\endgroup$
    – Derek Holt
    Jun 21, 2018 at 17:58
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    $\begingroup$ @Izzhov Google for "gnu moa group". The first hit is a paper according to which for every $n\le 10^7$, there exists $m$ with $gnu(m)=n$ $\endgroup$
    – Peter
    Jun 24, 2018 at 12:37

1 Answer 1

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That is not known (as far as I am aware). There is some relevant discussion in the book by Blackburn, Neumann, and Vekataraman "Enumeration of finite groups". The relevant section is $21.6$ "Surjectivity of the enumeration function" on page $238$.

While I have not read through that section, my understanding is that the authors do not provide a definite answer there. (Though they point out that this question has been asked before, several times; see below.) My suspicion is confirmed by the fact that it is repeated as an open problem on page $268$ (Question $22.36$).

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